This relation says that a self-adjoint, bounded, linear operator is greater than another operator of the same category if and only if the inner product of that operator with an arbitrary element in the complex Hilbert space is larger.

So to prove 1. we have to show that [itex]T \leq T[/itex] if and only if [itex]\langle Tx\,|\,x \rangle \leq \langle Tx\,|\,x \rangle[/itex]. Well, quite obviously, [itex]\langle Tx\,|\,x \rangle = \langle Tx\,|\,x \rangle[/itex], for any choice of [itex]x\in\mathcal{H}[/itex] and for any S-A,B,L operator on complex Hilbert space. Therefore (1) is proved.

Looks good. I wouldn't say, "since <.|.> is a number..." since technically it is just a complex number, and in general, <Tx|x> need not be a real number, and hence saying <Tx|x> < <Sx|x> may not make sense if <Tx|x> are complex (not purely real) numbers. There's no need to talk about what kind of number <Tx|x> is (although, is there a theorem that says it is real for self-adjoint operators?), just use the fact that < is a partial-order (in fact, a total order on the reals but meaningless in general on C) to conclude that if <Tx|x> < <Sx|x> and vice versa, that <Tx|x> = <Sx|x> and then you do as you did, concluding that T = S.

But proving the final bit: that for all [itex]T[/itex] (self-adjoint, bounded, linear)...

[tex]-\|T\|I \leq T \leq \|T\|I[/tex]

...where [itex]I[/itex] is the identity operator on the complex Hilbert space.

Im not sure I understand this statement. Is it saying that [itex]T[/itex] is bounded above and below? That is, let [itex]m=-I[/itex] and [itex]M=I[/itex], then this says that [itex]T[/itex] is an isomorphism no?

Is this a correct understanding of the last statement? I want to try to understand it before I prove it.

Just jumping back a second...you said "may not make sense if <Tx|x> are complex (not purely real) numbers.".

Then the statement "...since [itex]\leq[/itex] is a partial order on [itex]\mathbb{R}[/itex] (but meaningless in [itex]\mathbb{C}[/itex]) we can conclude that [itex]\langle Tx\,|\,x \rangle = \langle Sx\,|\,x \rangle[/itex]..." doesn't work because we are considering a complex Hilbert space so the inner products can be complex..wait...wait....

OMG...

I see what you mean now...aha! Self-adjoint operators only have REAL inner products! and we are considering only self-adjoint, bounded, linear operators. Yes, so [itex]\leq[/itex] is actually a partial order. Great.

I was wondering, to prove this part (the [itex]-\|T\|I \leq T \leq \|T\|I[/itex] for all S-A,B,L operators bit) will I need to use the following fact:

If you have two normed linear spaces X and Y and you also have a surjective linear map T such that [itex]T:X\rightarrow Y[/itex]. Then T is an isomorphism if and only if there exist [itex]m,M > 0[/itex] such that

I want to try to prove the given two normed vector spaces, X and Y, and a surjective linear map, T:X->Y, then

T is an isomorphism <=> there exist m,M > 0 such that

[tex]m\|x\| \leq \|Tx\| \leq M\|x\| \quad \forall \, x \in X[/itex]

But all I could come up with is this:

[tex](\Rightarrow)[/tex]
Suppose [itex]T[/itex] is an isomorphism. Then [itex]T[/itex] is continuous. Take [itex]\epsilon > 0[/itex] and in particular [itex]\epsilon = 1[/itex]. Then we have a [itex]\delta > 0[/itex] such that

Something is definitely wrong, where you said that you take m = 1. Let T(x) = x/2, you'll see you can't take m = 1. Here's one thing that's useful. If there is an m > 0 such that m|x| < |Tx| for all x, then the kernel of T must be 0, so T is injective. Given that it's a surjective linear map, it's an isomorphism. Conversely, assume that T is an isomorphism, hence continuous. I don't know what kind of theorems you have at your disposal and whether you can make use of notions like compactness, but if you can, consider the unit sphere in X, that is {x in X : |x| = 1} = S. The function R : X --> R defined by R(x) = |T(x)| is a composition of continuous functions, hence it's continuous and real-valued. Since S is compact, then a generalized version of the extreme value theorem tells you that R(S) achieves a maximum and minimum value. These values will naturally be positive since |.| is positive, and will be non-zero because the minimum value is non-zero, since if it were, then T(x) = 0 for some non-zero x, but T is an isomorphism, so this wouldn't be the case. Even if you can't use some of these theorems and ideas, it might be a good way to think about it.

Look at the set S. If R were unbounded above, then it would be that as you approach some s on S, R increases. But R must take on some fixed value at s itself, so it can't increase without bound.

AKG, so you are saying that I can prove boundedness of a surjective linear operator by considering a compact unit sphere?

No, I'm saying that you can prove the boundedness of a continuous function by considering a compact domain. In this case, the compact domain happens to be the unit sphere. Note that the function R defined by R(x) = |Tx| is not a linear operator. Use the fact that T is linear to conclude that it is continuous. Use the fact that absolute value is continuous, and that R is absolute value composed with T, to conclude R is continuous is continuous. Use the fact that S is compact to deduce R achieves its minimum and maximum values on S. Use the fact that T is injective (since it's an isomorphism) and the fact that |x| = 0 iff x = 0 to conclude that R(S) does not contain 0. Use the fact that |x| > 0 for all x to conclude that R(S) is contained in the positive reals. Let m be the minimum value of R, and M be the maximum value of R, on S.

When I say unit sphere, I mean {x in V (or whatever your vector space is) | |x| = 1}. It doesn't matter what the norm is, I'm calling it a sphere anyways. It might end up looking like something else, so if you want, you can say that this is the sphere with respect to this norm.

For a function to be bounded, that generally means bounded above and bounded below. However, all you need to know is that R achieves its minimum and maximum values. What you will be able to conclude is that for all x on the sphere:

m < R(x) < M
m|x| < |Tx| < M|w|

Let v be any vector now. v = |v|w, for some w on the unit sphere (w is the unit vector of v, like v-hat). Given the above inequalities, you can say:

m|w| < |Tw| < M|w|

since w is on the unit sphere. Multiply the whole thing by |v| and you're done.

Remember, a linear operator is a linear mapping from V to V. A linear transformation is more generally a linear mapping from V to W, vector spaces (so if in particular W = V, we call the transformation an operator). The function that sends x to |Tx| is not even linear, and certainly not an operator. Linear transformations, I believe, are always continuous, but you may want to check that. At very least, you have that T is an isomorphism, and hopefully that will be good enough to give you continuity.

Now I don't know about T continuous <=> |T| continuous. It may be true, but only because linear functions are always continuous, absolute value is always continuous, and the composistion of continuous functions are always continuous. But you can conceivably have a discontinous function T such that |T| is continuous, e.g. T = 1 for x < 0, T = -1 otherwise.

Firstly, it seems like you are using the extreme value theorem which says that a continuous function defined on some closed interval attains its minimum and maximum.

I want to know how you made that connection. I mean, I was talking about surjective linear maps between normed vector spaces, and somehow you come up with a method which turns |Tx| into a composition of continuous functions (absolute value and T itself) then let R be that function.

Also, am I correct in saying that to use this theorem we need a closed interval, and you have come up with an equivalent compact sphere, which I understand does the same thing.

So all I am saying is, how did you know to convert this problem of using operators into functions and then use the Extreme value property. How did you do it man!?

The domain has to be compact (not just closed, and doesn't have to be an interval of any sort). Specifically, if f : X --> R is a continuous function and X is a compact space, then f achieves its minimum and maximum. To prove that our domain, S, is compact, it suffices to show that it is closed and bounded. It is bounded because we defined it to have all its elements with norm 1, and {1} is certainly a bounded set. It is closed since its complement is open. How do we know? Well, pick any element not in S, say x with norm 1 + e, where e > 0. Choose an open ball (where the "ball" may look like a cube, or diamond, etc. based on the metric) around x with radius e/2. Suppose there is some y in this ball with norm 1. Then:

|x| = |(x-y) + y| < |x-y| + |y| < e/2 + 1 < 1 + e, contradiction.

You can show a similar thing for e < 0. So every point not in S is contained in an open ball that doesn't intersect S, so S-complement is open, so S is closed. When our domain is a subset of R, then the compact domains are the closed and bounded subsets of the reals, which are unions of finitely many closed intervals [a,b]. But the theorem I used is the generalized extreme value theorem, which works for any compact domain.

How did I know to do this? I'm not sure I can think of a general reason, since I would use this method only when the problem is specifically like this, i.e. to show that there is some M such that |Tx| < M|x| for all x. One reason is that on the unit sphere, |x| = 1. Since T is linear, you can know what T does to any vector v = |v|w, where w is the unit vector of v, v-hat, just by looking at Tw, and multiplying by |v|.